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13.02.2006 FINITE ELEMENT FORMULATION

Chapter 2FINITE ELEMENT FORMULATION

Zahit Mecitoğlu

FINITE ELEMENT ANALYSIS IN STRUCTURES

© 2006 İstanbul Technical University

FINITE ELEMENT FORMULATION13.02.2006

AIMThe basic aim of solid mechanics problem is to determinethe distribution of displacements and stresses under the loading and boundary conditions.

A mathematical model of the structural problem is necessary to find the desired distributions by using the FEM.

Hence, the basic equations of solid mechanics are summarized in the following sections for ready reference in the formulation of FE equations.


BASIC EQUATIONS

For a reasonable solution of a problem, following basicequations must be satisfied for an elastic continua.

Strain-displacement relationsConstitutive relationsEquilibrium equationsCompatibility equationsBoundary conditions


FIELD VARIABLESThe loadings and boundary conditions on an elastic body cause the displacements, strains and stresses inside the body.

=PA

σ

P Pσ

L u=

uL

ε

A


Strain-Displacement RelationsThe deformed shape of an elastic body under loads can be completely described by the three components of displacement u, v, and w.

The strains induced in the body can be expressed in terms of the displacement components.

Geometric linearity : deformations are small enough so that the strain-displacement relations remain linear.


...Strain-Displacement Relations

0 0

0 0

0 0

0

0

0

∂⎡ ⎤⎢ ⎥∂⎢ ⎥

∂⎢ ⎥⎧ ⎫ ⎢ ⎥∂⎪ ⎪ ⎢ ⎥∂⎪ ⎪ ⎢ ⎥ ⎧ ⎫⎪ ⎪ ⎢ ⎥∂⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎢ ⎥∂ ∂⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎩ ⎭∂ ∂⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥∂ ∂⎪ ⎪ ⎢ ⎥⎩ ⎭

∂ ∂⎢ ⎥⎢ ⎥∂ ∂⎢ ⎥

∂ ∂⎣ ⎦

x

y

z

xy

yz

zx

x

y

uz v

wy x

z y

z x

εε

εγ

γ

γ

{ } [ ]{ }= dε δ

For the 3-D case, the strain-displacement relations can be written as follows,


In the case of linearly elastic isotropic three-dimensional solid, the stress-strain relations are given by Hooke’s law.

( ) ( ) ( )( )

( )

12

12

12

1 0 0 01 0 0 0

1 0 0 00 0 0 1 2 0 01 1 20 0 0 0 1 2 0

0 0 0 0 0 1 2

−⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥

−⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬−+ − ⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪−⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎣ ⎦

x x

y y

z z

xy xy

yz yz

zx zx

E

σ εν ν νσ εν ν ν

ν ν νσ εντ γν ν

ντ γντ γ

{ } [ ]{ }= Eσ ε

Constitutive relations


Equilibrium EquationsIf we consider an element of material inside the body, it must be in equilibrium due to the internal stresses developed by the loads.

y

x

z

σy

σx

σz

τyx

τxyτyz

τzy

τzxτxz

x

Φ

P1

P2

y

z

(a)

Sσ

S


…Equilibrium EquationsTheoretically the state of stress at any point of a body is completely defined in terms of the nine components of stress.

The normal stresses are σx, σy, and σz.

Shear stresses are τxy, τyx, τyz, τzy, τxz, and τzx.


…Equilibrium Equations

0

0

0

∂∂ ∂+ + + =

∂ ∂ ∂∂ ∂ ∂

+ + + =∂ ∂ ∂

∂∂ ∂+ + + =

∂ ∂ ∂

xyx xzx

yx y yzy

zyzx zz

bx y z

bx y z

bx y z

τσ τ

τ σ τ

ττ σ

The moment equilibriums about the x, y and z axis give the relations

= = =xy yx xz zx yz zyτ τ τ τ τ τ

The force equilibrium in x, y and z directions give the following equilibrium equations

bx, by , bz : body forces per unit volume.


…Equilibrium Equations

0 0 00

0 0 0 00

0 0 0

⎧ ⎫⎡ ⎤∂ ∂ ∂ ⎪ ⎪⎢ ⎥∂ ∂ ∂ ⎪ ⎪ ⎧ ⎫⎢ ⎥ ⎧ ⎫⎪ ⎪∂ ∂ ∂ ⎪ ⎪⎢ ⎥ ⎪ ⎪ ⎪ ⎪− =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥∂ ∂ ∂ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎩ ⎭⎩ ⎭⎪ ⎪∂ ∂ ∂⎢ ⎥⎪ ⎪⎢ ⎥∂ ∂ ∂⎣ ⎦ ⎪ ⎪⎩ ⎭

x

yx

zy

xyz

yz

xz

x y z bb

y x zb

z y x

σσ

στ

τ

τ

In a compact format

[ ] { } { } { }0− =Td bσ

In matrix format we may write


Many problems in theory of elasticity are two dimensional in nature and they can be modeled as plane stress and plane strain.

Plane Stress : A thin plate subjected to in-plane loading acting in its own plane, the state of stress and deformation within the plate is called plane stress. In this case only two dimensions (in the plane of plate) are required for the analysis.

SPECIAL CASES: 2-D CASE


PLANE STRESS

x

y

z

x

y

A thin plate under inplane loading.

...SPECIAL CASES: 2-D CASE


Plane Strain: If a long body is subjected to transverse loading and its cross section and loading do not vary significantly in the longitudinal direction, a small thickness in the loaded area can be treated as subjected to plane strain.



PLANE STRAIN

A long cylinder under internal pressure.

z

x

y

z

y

p



…SPECIAL CASES: 2-D CASEFor a two-dimensional problem, there will be only three independent stress components (σx, σy, τxy) and three independent strain components (εx, εy, γxy)

0

0

∂∂+ + =

∂ ∂∂ ∂

+ + =∂ ∂

xyxx

xy yy

bx y

bx y

τσ

τ σ

strain-displacement relations0

0

⎡ ⎤∂⎢ ⎥∂⎧ ⎫ ⎢ ⎥

⎪ ⎪ ⎧ ⎫∂⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥∂ ⎩ ⎭⎪ ⎪ ⎢ ⎥⎩ ⎭ ∂ ∂⎢ ⎥

⎢ ⎥∂ ∂⎣ ⎦

x

y

xy

xuvy

y x

εε

γ

equilibrium equations


The constitutive equations for plane stress case

0= = =z zx yzσ τ τ ( ) 0= − + = =z x y yz zxEνε σ σ ε ε

( )2

1 01 0

1 10 02

⎡ ⎤⎧ ⎫ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥−⎩ ⎭ ⎩ ⎭⎢ ⎥⎣ ⎦

x x

y y

xy xy

Eσ ν εσ ν ε

ν ντ γ



w = 0 and at every cross section.0∂=

∂wz

0= = =z zx yzε γ γ ( ) 0= + = =z x y yz zxσ ν σ σ τ τ

( ) ( )

1 01 0

1 1 21 20 0

2

⎡ ⎤⎧ ⎫ ⎧ ⎫⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥= −⎨ ⎬ ⎨ ⎬⎢ ⎥+ −⎪ ⎪ ⎪ ⎪⎢ ⎥−⎩ ⎭ ⎩ ⎭⎢ ⎥⎣ ⎦

x x

y y

xy xy

Eσ ν ν εσ ν ν ε

ν νντ γ

The constitutive equations for plane strain case



…SPECIAL CASES: 1-D CASE

Equilibrium equation 0+ =xx

d bdxσ

Constitutive Relation1or= =x x x xEE

σ ε ε σ

Strain-Displacement Relation =xdudx

ε

In one-dimensional problems, only one component of stress, namely, σx will be there.


The Number of Basic EquationsType of equations Number of equations

In 3-D problems In 2-D problems In 1-D problems

Strain-Displacement Rel.

6 3 1

Constitutive Relations 6 3 1

Equilibrium Equations 3 2 1

Total Number of Eqs. 15 8 3


The Number of UnknownsUnknowns In 3-D problems In 2-D problems In 1-D problems

Displacements u, v, w u, v u

Strains εx, εy, εz, γxy, γyz, γzx εx, εy, γxy εx

Stresses σx, σy, σz, τxy, τyz, τzx σx, σy, τxy σx

Total number of unknowns

15 8 3


Summary

[ ]{ } { } { }0Δ − =bσ

{ } [ ]{ }= Eσ ε

{ } [ ]{ }= dε δ

Equilibrium equations

Constitutive relations

Strain-displacement relations


POTENTIAL ENERGY

The potential energy of an elastic body Π is defined asΠ = U - W

In this equation U is the strain energy and W is the work done on the body by the external forces.

The strain energy of a linear elastic body is defined as

{ } { } { } [ ]{ }1 12 2

= =∫ ∫T T

V V

U dV E dVε σ ε ε


…POTENTIAL ENERGYThe work done by the external forces can be stated as

{ } { } { } { } { } { }1

11=

= − Φ + ∑∫ ∫I

T T Ti i

iV S

W b dV dS Pδ δ δ

{b} is the vector of body forces.{Φ} is the vector of distributed surface force.{P} is the vector of concentrated loads.


FORMULATIONS OF SOLID AND STRUCTURAL MECHANICS

Most continuum problems can be formulated according to one of the two methods.

differential equation methodvariational method


...FORMULATION METHODSFormulation Methods

Differential equation formulation methods

Variational formulation methods

Displacement method

Force method

Displacement-force method

Principle of minimum potential energy

Principle of minimum complementary energy

Principle of stationary Reissner energy


MINIMUM POTENTIAL ENERGY

Most of the equations in this course will be derived from the potential energy theory.

We assume that strains and displacements are small and no energy dissipated in the static loading process.

That is, the external work of gradually applied loads is equal to the energy stored in the structure, and the system is said to be conservative.


MINIMUM POTENTIAL ENERGY

The principle says that among all the displacement states of a conservative system that satisfy compatibility and boundary restraints, those that also satisfy equilibrium make the potential energy stationary.

δΠ = δU - δW = 0

Variation is taken with respect to the displacements.


F.E. FORMULATION

As a first step, the displacement field of an element is approximated by using interpolation functions and nodal displacements.

{ } [ ]{ }( , , )( , , )( , , )

⎧ ⎫⎪ ⎪= =⎨ ⎬⎪ ⎪⎩ ⎭

u x y zv x y z N qw x y z

δ e

x, u

y, v

q1

q2

q3 q5

q6q4

1

2

3

z, w

4

q7

q8

q9

q10

q11

q12


… F.E. FORMULATION

Strain vector can be expressed in terms of nodal displacements {q}, by using strain-displacement relations

{ } [ ]{ } [ ][ ]{ } [ ]{ }== =d d BN q qε δ

{ } [ ]{ } [ ][ ]{ }= = BE qEσ ε

Stress vector can be expresses in terms of nodal displacements, {q}, by using the constitutive relations.


… F.E. FORMULATIONThe strain energy of an element can be written in terms of nodal displacements as

{ } [ ]{ }

{ } [ ] [ ] [ ]{ }

{ } [ ] [ ] [ ] { }

{ } [ ]{ }

12

12

12

12

=

=

=

=

∫

∫

∫

e

e

e

T

T T

e

V

VT

T

T

V

U E dV

E dV

q q

q B B q

B E B dV

q qk

ε ε



{ } { } { } { }

{ } [ ] { } { } [ ] { }

{ } [ ] { } { } [ ] { }

{ } { } { } { }

1

1

1

1

1

1

= + Φ

= + Φ

= Φ+

= +

∫

∫

∫ ∫

∫

∫

e e

e e

e e

e

V S

V

T T

T TT T

T T

V S

b s

S

T T

T T

W b dV dS

b dV dSq N q N

N b dV N dS

q f

q q

q f

δ δ

The work done by the external forces The work done by the external forces for for an element can be written in terms of nodal displacements as



{ } [ ] { } Element body force vector= ∫e

Tb

V

f N b dV

If an element has not a boundary at the surface under distributed load, the contribution of surface forces will be zero for this element.

{ } [ ] { }1

Element surface load vector= Φ∫e

Ts

S

f N dS

[ ] [ ] [ ] [ ] Element stiffness matrix= ∫e

T

V

k B E B dV

The formulae for the element stiffeness and load vector remain the same irrespective of the type of the element. However, the order of the stiffness matrix and the load vector will change for different types of elements.


… F.E. FORMULATIONThe potential energy of an element can be written as

{ } { }1=

Π = Π −∑E

T

ec

e FQ

{ }

1

2

1=

⎧ ⎫⎪ ⎪⎪ ⎪= =⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

∑E

e

N

QQ

Q q

Q

{Q} is the vector of nodal displacements of the entire structure

Microsoft Word Document

The potential energy of the entire structure can be obtained from the summation of the element energies.

{ } [ ]{ } { } { } { } { }12Π = + = − −T T Te e e

b skW fU q q q fq


… F.E. FORMULATIONASSEMBLY: The summation of the equation implies the expansion of element matrices to “structure size” followed by summation of overlapping elements.

{ } [ ]{ } { } { } { } { } { } { }

{ } [ ]{ } { } { } { } { }( )

12

1 1 1

12

= = =Π = −

+

− −

= − +

∑ ∑ ∑E E E

T T T T

e e eb s c

b sT

cT

q q q q Q

Q Q

k f f F

K F F FQ

{ } [ ]{ } { } { }12Π = −T TKQ Q Q F



[ ] [ ]1=

= ∑E

eK k

The element stiffness matrix and the global stiffness The element stiffness matrix and the global stiffness matrix are always symmetric.matrix are always symmetric.

Some of the contributions to the load vector {F} may be zero in a particular problem.

Some of the components of the load vectors may be moments if the corresponding nodal displacements represent rotations.

Assembly procedure gives us the global stiffeness matrix and load vector.

{ } { } { }( ) { }1=

= + +∑E

b s ce

F f f F



0 1,2,3,Π= = …

ii N

Q∂∂

[ ]{ } { }=K Q FThe required solution for the nodal displacements and element stresses can be obtained after solving above eqs.

The linear algebraic equations of the overall structure can now be expressed as

The static equilibrium equations of the structure can be obtained from the principle of minimum potential energy.


The equilibrium equation, [K]{Q}={F}, cannot be solved since the stiffness matrices [k] and [K] are singular, and hence their inverses do not exist.

The physical significance of this is that a loaded structure is free to undergo unlimited rigid body motion (translation and/or rotation) unless some support or boundary constraints are imposed on the structure to suppress the rigid body motion.

These constraints are called boundary conditions.

We solve the equations after incorporating the prescribed boundary conditions to obtain the displacements.



INITIAL STRAINS

{ } { }( )0= −eε ε ε

If some initial strains, such as temperature changes, exist in the structure, the elastic portion of the total strain vector can bestated as

{ } [ ] { } { }( )0= −Eσ ε εand stress vector is writen as

P

ΔTΔT


...INITIAL STRAINS

{ } { }( ) [ ] { } { }( )

{ } [ ]{ } { } [ ]{ } { } [ ]{ }

0 0

0 0 0

12

1 12 2

= − −

= − −

∫

∫ ∫ ∫e e e

Te

V

T T T

V V V

U E dV

E dV E dV E dV

ε ε ε ε

ε ε ε ε ε ε

The strain energy of an element can be found

The first term yield the element stiffness matrix. Since the last term is constant, its contribution is zero during the minimization of the total potential energy.


...INITIAL STRAINS

{ } [ ] [ ]{ }0= ∫e

Tt

V

f N E dVε

{ } [ ]{ }

{ } [ ] [ ]{ }

{ } [ ] [ ]{ }

{ } { }

0

0

0

= −

= −

= −

= −

∫

∫

∫e

e

e

Tet

VT T

VT

VT

T

t

W E dV

q N E dV

q

N E Vq d

f

ε

ε ε

ε

Element nodal force vector due to initial strains

The second term yields the element nodal force vector due to initial strains.

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